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Let $A$ be a commutative ring, prove that if $x \in A$ is nilpotent then $1-x$ is an invertible element in $A$.

I need help with this one.

Mikasa
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2 Answers2

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Let's do it "informally" first to get an intuition:

$(1-x)^{-1}$ "should be" = $1+x+x^2+...x^{n}+...$. Now since $x$ is nilpotent, there exists a (minimal) $n$ such that $x^{n+1}=0$ but $x^n \neq 0$. So the above series reduces to a finite sum $1+x+...x^n$.

Now back to the formal world. Due to the insight gained above, we would conjecture that $1+x+...x^n$ is an inverse for $(1-x)$. Check that this indeed is the case by direct multiplication.

voldemort
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3 solutions :

a) If $x$ is nilpotent, $x \in N \subset J$, where $N$ is the nilradical and $J$ the Jacobson radical. And another characterisation of the Jacobson radical is precisely : $x \in J \Leftrightarrow \forall y \in A, 1 - xy$ is a unit.

b) Since $(1-x)(1+x) = (1-x^2)$, $(1-x^2)(1+x^2) = (1-x^4)$, ..., there is a rank $k$ such that $2^k > n$. then, $(1-x)(1+x)(1+x^2)...(1+x^{2^{k-1}}) = (1-x^{2^{k}}) = 1$.

c) Use the geometric serie.